# Acid base reations

1. Jan 2, 2010

### hallowon

1. The problem statement, all variables and given/known data
after titrating sodium hydroxide with hydrofluoric acid, a chemist determined that the reaction had formed an aqueous solution of 0.020 mol/l sodium fluoride. determine ph of the solution

2. Relevant equations

3. The attempt at a solution
Ka of HF= 6.3X10^-4
NaOH + HF (equlibrium arrows) NaF +H20
I ? ? 0 0
C -X -X +X +X
E ?-X ?-X 0.020 0.020

ka = product/reactants i got stuckhere because there no initial value, and i don't know why im trying to find initial when i need to find ph

2. Jan 2, 2010

### Staff: Mentor

You look for pH of 0.020M F- solution, everything else doesn't matter. This is a weak base.

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3. Jan 2, 2010

### hallowon

wait i don't get it ? -log 0.020M is 1.69 that is wrong b/c Strong base reacts with weak acid so ph should be aboe 7 b/c it will be a basic solution

4. Jan 2, 2010

### x12179x

You don't need to know I, the initial, since we are given that the Equilibrium of NaF (C-X) is 0.020

The dissociation equation, as you said is:

NaF + H20 <==> NaOH + HF
(B + H2O <==> HB + OH-)

or:

F- + H2O <=> HF + OH-

I C ........N/A.............
C C -X .......... +X .... +X
E 0.020 .......... +X .... +X

Since we have B + H2O <==> HB + OH-, the equation to use is then Kb=[HB][OH-]/[B-] so we need to get Kb from Ka so...

Kb = Kw/Ka = [10^-14]/[6.3x10^-4] = 1.58 x10^-11​

We now know:
Kb=1.58 x10^-11
[F-]=0.02
[OH-]=[HF]=X

All you have to do is plug these values into Kb=[HF][OH-]/[F-] and solve for X
Remember... X=[OH-], so you need to get the pH from the pOH

Relevant equations:
B + H2O <==> HB + OH-
Kb = Kw/Ka
Kb=[HB][OH-]/[B-]
pOH= -log(OH-)
pOH + pH=14

Last edited: Jan 3, 2010
5. Jan 3, 2010

### hallowon

thanks so much!! you just made my day.

6. Jan 3, 2010

### Staff: Mentor

No, we are given that initial F- is 0.020M, and equilibrium concentration of HF is not 0.020M but has to be calculated. And in fact it is a lot smaller than 0.020M.

You will be badly surprsised if you assume x12179x approach.

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7. Jan 3, 2010

### x12179x

Hmm, I mistyped HF instead of NaF above (and also .2 instead of .02); Now corrected

But I thought 0.02 was the equilibrium value of NaF since it said:

"after titrating...the reaction had formed an aqueous solution of 0.020 mol/l sodium fluoride." ?

8. Jan 3, 2010

### hallowon

thats kinda what i assumed to

9. Jan 4, 2010

### Staff: Mentor

Just because you have 0.020M solution of NaF doesn't mean 0.020M is an equilibrium concentration of F-. Quite the opposite - F- is a weak Broensted base, reacting with water, so its concentration after hydrolysis is smaller.

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